Question

If the function is bijective, is the inverse also bijective ?

Answer 1

for f(x) :
its surjective, that gives me range = codomain
its injective, that gives me one-to-one

Answer 2

i can see that the inverse is injective
what can i say about the surjectiveness ?

Answer 3

Let \(f:X\rightarrow Y\) be a bijective function. Then the inverse is well defined and we denote it by \(f^{-1}:Y\rightarrow X\) (furthemore, we note that \(f\circ f^{-1} = f^{-1}\circ f=\mathrm{id}\)). Since we know that \(f(X)=Y\) (because \(f\) is surjective), then \(f^{-1}(f(X)) = f^{-1}(Y) \implies f^{-1}(Y)=X\). Hence \(f^{-1}:Y\rightarrow X\) is surjective.

Does this make sense? :-)

Answer 4

surjective implies range being filled with the entire codomain
however there could be few points in domain at which the function is not defined, right ?